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  4. By Simpson rule taking n=4, the value of the integral ∫01 (1/1+x2) d x is equal to

Q. By Simpson rule taking $n=4$, the value of the integral $\int_{0}^{1} \frac{1}{1+x^{2}} d x$ is equal to

MHT CETMHT CET 2011 Report Error

Solution:

Here, $h=1 / 4,=0.25, y=\frac{1}{1+x^{2}}$
x y
1 0 1.0
2 0.25 0.941
3 0.5 0.8
4 0.75 0.64
5 1 0.5

By Simpson's Rule
$\int_\limits{0}^{1} \frac{d x}{1+x^{2}}=\frac{1}{4 \times 3}$
$[(1+0.5) + 4(0.941+ 0.64)+2(0.8)]$
$=\frac{1}{12}[9,424]=0.785$

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